Question

If $ \alpha ,\beta $ and $ \gamma $ are the roots of equation $ {{x}^{3}}-3{{x}^{2}}+x+5=0, $ then $ y=\Sigma {{\alpha }^{2}}+\alpha \beta \gamma $ satisfies the equation.

• A. $ {{y}^{3}}+y+2=0 $
• B. $ {{y}^{3}}-{{y}^{2}}-y-2=0 $
• C. $ {{y}^{3}}+3{{y}^{2}}-y-3=0 $
• D. $ {{y}^{2}}+4{{y}^{2}}+5y+20=0 $

Answer 1

Answer: (B)

Given $ \alpha ,\beta ,\gamma $ are the roots of equation
$ {{x}^{3}}-3{{x}^{2}}+x+5=0 $
$ \therefore $ $ \alpha +\beta +\gamma =3 $
$ \alpha \beta +\beta \gamma +\gamma \alpha =1 $
and $ \alpha \beta \gamma =-5 $
Now, $ y=\Sigma {{\alpha }^{2}}+\alpha \beta \gamma $
$ ={{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}+\alpha \beta \gamma $
$ ={{(\alpha +\beta +\gamma )}^{2}}-2(\alpha \beta +\beta \gamma +\gamma \alpha )+\alpha \beta \gamma $
$ ={{(3)}^{2}}-2(1)-5 $
$ \Rightarrow $ $ y=2 $
So, $ y=2 $
satisfies the equation $ {{y}^{3}}-{{y}^{2}}-y-2=0 $